1. Proof of Lemma 3.1

Lemma 3.1
Suppose that \(G=\langle N, \{S_i, u_i\}_{i\in N}\rangle\) is a
strategic game. A strategy \(s_i\in S_i\) is strictly dominated
(possibly by a mixed strategy) with respect to \(X\subseteq S_{-i}\) iff
there is no probability measure \(p\in \Delta(X)\) such that \(s_i\) is a
best response with respect to \(p\).

Proof: We start with some preliminary observations. Let
\(G=\langle S_1, S_2, u_1, u_2\rangle \) be a two-player strategic
game. Recall that \(\Delta(S_1)\) and \(\Delta(S_2)\) denote the mixed
strategies for players 1 and 2, respectively. For \(p_1\in
\Delta(S_1)\), \(p_2\in \Delta(S_2)\), we write \(U_1(p_1, p_2)\)
(respectively \(U_2(p_1,p_2)\)) for the expected utility that player \(1\)
(respectively \(2\)) receives when \(1\) uses the mixed strategy \(p_1\) and
\(2\) uses the mixed strategy \(p_2\). We assume that the players’ choices
are independent, so we have the following calculation for \(i=1,2\):

A two-player game \(G=\langle S_1, S_2, u_1, u_2\rangle \)
is zero-sum provided for each \(x\in S_1\) and \(y\in
S_2\), \(u_1(x,y)+u_2(x,y)=0\). We make use of the following fundamental
theorem of von Neumann:

Theorem S2 (von Neumann’s minimax theorem) For
every two-player zero- sum game with finite strategy sets \(S_1\) and
\(S_2\), there is a number \(v\), called the value of the
game such that:

Now, we can proceed with the proof of the Lemma. Suppose that
\(G=\langle N, \{S_i, u_i\}_{i\in N}\rangle\) is a strategic game
where each \(S_i\) is finite.

Suppose that \(s_i\in S_i\) is strictly dominated with respect to
\(X\). Then there is a \(s_i \in S_i\) such that for all \(s_{-i}\in
X\), \(u_i(s_i',s_{-i}) > u_i(s_i,s_{-i})\). Let \(p\in\Delta(X)\)
be any probability measure. Then for all \(s_{-i}\in X\), \(0\le
p(s_{-i})\le 1\). This means that for all \(s_{-i}\in X\), we have
\(p(s_{-i})\cdot u_i(s_i',s_{-i}) \ge p(s_{-i})\cdot
u_i(s_i,s_{-i})\), and there is at least one \(s_{-i}\in S_{-i}\) such
that

\[p(s_{-i})\cdot u_i(s_i',s_{-i}) > p(s_{-i})\cdot u_i(s_i,s_{-i})\]

(this follows since \(p\) is a probability measure on \(X\), so cannot
assign probability 0 to all elements of \(X\)). Hence,

So, \(EU(s_i',p)> EU(s_i,p)\), which means \(s_i\) is not a
best response to \(p\).

For the converse direction, we sketch the proof for two player
games. The proof of the more general statement uses the supporting
hyperplane theorem from convex analysis. We do not discuss this
extension here (note that it is not completely trivial to extend this
result to the many agent case as we must allow players to have beliefs
about correlated choices of their opponents). Let \(G=\langle
S_1, S_2, u_1, u_2\rangle\) be a two-player game. Suppose that
\(\alpha\in \Delta(S_1)\) is not a best response to any
\(p\in\Delta(S_2)\). This means that for each \(p\in \Delta(S_2)\)
there is a \(q\in\Delta(S_1)\) such that
\(U_1(q,p)>U_1(\alpha,p)\). We can define a function
\(b:\Delta(S_2)\rightarrow \Delta(S_1)\) where, for each
\(p\in\Delta(S_2)\), \(U_1(b(p),p)>U_1(\alpha,p)\).

Consider the game \((S_1, S_2, \ou_1, \ou_2)\) where

\[\ou_1(s_1, s_2)=u_1(s_1,s_2) - U_1(\alpha, s_2)\]

and \(\ou_2(s_1,s_2)=-\ou_1(s_1,s_2)\). This is a 2-person,
zero-sum game, and so by the von Neumann minimax theorem, there is a
mixed strategy Nash equilibrium \((p_1^*, p_2^*)\). Then, by the
minimax theorem, for all \(m\in\Delta(S_2)\), we have

Hence, for all \(m\in\Delta(S_2)\) we have \(\oU_1(s_1^*,m)>0\)
which implies for all \(m\in\Delta(S_2)\), \(U_1(s_1^*,m) >
U_1(\alpha, m)\), and so \(\alpha\) is strictly dominated by
\(p_1^*\). QED

2. Proof of Theorem 6.1

Theorem 6.1
(Brandenburger & Keisler 2006: Theorem 5.4) There is no
assumption-complete type structure for the set of conjectures that
contain the first-order definable subsets.

To prove this theorem, we follow an idea recently discussed in
Abramsky & Zvesper (2010). Suppose that \(\C_A\subseteq\pow(T_A)\)
is a set of conjectures about Ann states (similarly, let
\(\C_B\subseteq\pow(T_B)\) be a set of conjectures about Bob states). We
start with the flowing assumption:

Lemma S1
Under the above assumption, for each \(X\in \C_A\) there is an \(x_0\)
such that \[x_0\in X\text{ iff there is a \(y\in T_B\) such that \(y\in
\lambda_A(x_0)\) and \(x_0\in\lambda_B(y)\)}\]

Proof of Lemma S1:
Suppose that \(X\in\C_A\). Then there is an \(x_0\in T_A\) satisfying 1
and 2.

Suppose that \(x_0\in X\). By 1., \(\lambda_A(x_0)\ne\emptyset\) so
there is a \(y_0\in T_B\) such that \(y_0\in\lambda_A(x_0)\). We show
that \(x_0\in\lambda_B(y_0)\). By 2., we have
\(y_0\in\lambda_A(x_0)\subseteq \{y \mid \lambda_B(y)=X\}\). Hence,
\(x_0\in X=\lambda_B(y_0)\), as desired.

Suppose that there is a \(y_0\in T_B\) such that
\(y_0\in\lambda_A(x_0)\) and \(x_0\in \lambda_B(y_0)\). By 2.,
\(y_0\in\lambda_A(x_0)\subseteq \{y \mid \lambda_B(y)=X\}\). Hence,
\(x_0\in\lambda_B(y_0)=X\), as desired.

\(\neg\phi(x):=\forall y(R_A(x,y)\rightarrow\neg R_B(y,x))\): “all
states \(x\) where any state \(y\) that Ann considers possible is such
that Bob does not consider \(x\) possible at \(y\).” That is, this
formulas says that “Ann believes that Bob’s assumption
is wrong.”